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NCERT Exemplar · Q73

Q.The degree of the differential equation d2ydx2+(dydx)3+6y5=0\frac{d^2y}{dx^2}+\left(\frac{dy}{dx}\right)^3+6y^5=0 is:
(A) 1
(B) 2
(C) 3
(D) 5

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The order of this differential equation is 2 (highest derivative is d2ydx2\frac{d^2y}{dx^2}), and since the equation is already a polynomial in the derivatives with the highest-order derivative raised to the power 1, the degree is 1. The correct option is (A).

The degree of a differential equation is defined only when the equation is a polynomial in the derivatives. It is the power of the highest-order derivative present, after the equation has been made free of radicals and fractions in the derivatives.

Here, the highest derivative is d2ydx2\frac{d^2y}{dx^2} (order 2). The equation is:

d2ydx2+(dydx)3+6y5=0\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^3 + 6y^5 = 0

This is already a polynomial in d2ydx2\frac{d^2y}{dx^2} and dydx\frac{dy}{dx} — no square roots, no fractional powers of derivatives. The highest-order derivative d2ydx2\frac{d^2y}{dx^2} appears with exponent 1. So the degree is simply 1.

Watch out

A common mistake is to confuse the exponent on the lower-order derivative (dydx\frac{dy}{dx} is cubed here) with the degree. Degree is determined only by the exponent of the highest-order derivative. The cube on dydx\frac{dy}{dx} does not affect the degree. …

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